Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int N;
double p[3000],dp[3000][3000],ans;
int main()
{
    scanf("%d",&N);
    for(int i=1; i<=N; i++)
        scanf("%lf",&p[i]);
    dp[0][0]=1.0;
    for(int i=1; i<=N; i++)
    {
        dp[i][0]=dp[i-1][0]*(1-p[i]);
        for(int j=1; j<=i; j++)
            dp[i][j]=p[i]*dp[i-1][j-1]+(1-p[i])*dp[i-1][j];
    }
    for(int i=N; i>N/2; i--)
        ans+=dp[N][i];
    printf("%.9f\n",ans);
}